Calculating Water Evaporation in a Power Plant with Waste Heat Loss

[gsxrK3]

Homework Statement

An 800-MW electric power plant has an efficiency of 30%. It loses its waste heat in large cooling towers by evaporating a fine mist of water. If all waste heat went only into the evaporation of water, how many tons of water would be evaporated per day?

Homework Equations

W = |Qh|-|Qc|
e = 1 - (|Qc|/|Qh|)
Q = mL
Lwater = 2.26x10^6 J/kg

The Attempt at a Solution

I can't seem to get any of the answers listed and I can't quite figure out what to do. Here's what I tried.

W = 800MW
e = .30
Solving for |Qc| and |Qh| I got, |Qc| = 1866.69 MW, or 1866.69x10^6 J/s, |Qh| = 2666.67 MW
Since the wasted heat going to the cold reservoir is what is evaporating the water I plugged |Qc|= 1866.69x10^6 J/s into the Q for the equation Q = mL, so |Qc| = mL. L is the latent heat of water. Solving for m I got m = 825.97 kg/s. Since it's per day, I multiply m by 86400seconds and get 71,363.72 tons of water per day.

Would anybody like to take a crack at it and help me out? Thanks :)

 

[lalbatros]

gsxrK3,

Your solution seem correct to me.
I got a waste heat of 161280 GJ/day.
I like GigaJoules units because of my profession.
For the same reason I remember that the heat of evaporation of water is about 2.25 GJ/ton.
Therefore, your 71 ktons are ok.

Note that the latent heat depends of the water and vapor temperature.
But for the purpose here, it's fine.

Quite a lot of water, isn't it?

 

[gsxrK3]

Yeah that is a ton of water, I mean a lot. Much thanks for the reply. May I ask what it is you do?